//编程思路通过前缀和来处理
// 狂吃不止和线性出现
//  s[j] - s[i-1]     -> 一段区间美味值 
#include<bits/stdc++.h>
using namespace std;
int n, l, r;
int ans = 0;
const int N = 1e5+100;
int a[N];
int s[N];//前缀和 
int main()
{
freopen("bjoi2016_hzss.in","r",stdin);
freopen("bjoi2016_hzss.out","w",stdout);
cin >> n >> l >>r;    
for(int i = 1; i <= n; i++)
{
cin >> a[i];
s[i] = a[i] +s[i-1];
}
for(int i = 0; i <= n; i++)
{
for(int j = i+1; j <= n; j++)
{
if(s[j] - s[i] <= r && l <= s[j] - s[i])  
{
ans++;
}
}    
}
cout << ans;
} 
/*
6 7 17
7 5 1 4 2 8
*/